Decoherence and apparent collapse

[durant35]

By observe do you mean observe with our eyes or that the photons that hit eliminate the possibility of a superposition?

 

[stevendaryl]

By observe do you mean observe with our eyes or that the photons that hit eliminate the possibility of a superposition?

I meant something broader: Any experiment that you can think of to detect the effects of a macroscopic superposition will fail, if decoherence has taken place.

 

[durant35]

So we can conclude there are no macroscopic superpositions on a macroscopic scale? It seems bizarre and disturbing that any of us at the moment are for an instan both dead and alive like the cat.

 

[A. Neumaier]

And if I'm not observing the object during decoherence, what am I observing since air photons etc hit the object while I'm looking at it. For instance my cell phone. I see it is black, rectangular and so on. It seems that decoherence is present all the time.

As long as one uses a classical description, there is no decoherence. Decoherence only appears if you regard everything as a quantum system with as description so complicated that no one ever can give it, and assume that this description is 100% accurate to the last detail of every atom of your cell phone, that its antenna responds to every single photon in the electromagnetic wave created by your provider, and that you try to predict the response of a transistor in you microchip to this photon, based on statistical mechanics.

Nobody in his right mind is doing this. It is done only in extremely simplified model calculations to ensure that there is no contradiction between classical macroscopic observations and quantum mechanics - never in anything resembling complex reality. The talk about superpositions, decoherence, and all that is useful in these model calculations and model experiments, and only there. Apart from the context of these highly idealized models, it becomes misleading and mostly meaningless.

However, the same talk is transported to the general public by physicists and media who try to make their abstract work understandable by simplifying it beyond the recognizable, and it is magnified into horrible philosophical conundrums by those who want stories that sell well and appeal to the sense of mystery that many people like to nourish.

There are two sensible ways out:

• Either enjoy the mystery and be content with knowing that it is made up from something understandable in its proper context, but dissolved from it becomes a kind of science fiction.
• Or, learn the basics properly, which may mean to learn enough math and physics to understand the formulas and problems behind it, and to see how the talk is related to the formulas (and hence to the true physics).

But trying to grapple with these problems on the level of expositions for the general science-interested person will almost surely never lead out of the maze.

 

[kith]

in that thread Atyy wrote:
"In an improper mixed state, each system is in the same pure state."

The system he refers to is the combined system, which is an a pure, entangled state. The "measurement" he talks about is that of a hypothetical observable of the combined system which has the pure, entangled state as one of its eigenstates.

Now atyy apparently uses improper and proper synonymous with pure and mixed, while many others - e.g., http://arxiv.org/pdf/quant-ph/0109146.pdf - use improper for a state that is prepared by mixing pure states, and proper for a state that appears as a restriction of a pure state of a bigger system to a subsystem (which usually results in a mixed state).

[bolding mine]
You have mixed up -no pun intended ;-)- improper and proper here, the author uses them the other way round. Apart from that, my impression from several discussions with atyy is that he has the same definitions in mind (see above).

 

[stevendaryl]

So we can conclude there are no macroscopic superpositions on a macroscopic scale? It seems bizarre and disturbing that any of us at the moment are for an instan both dead and alive like the cat.

Yeah, for all practical purposes, it is impossible to have (for more than a tiny fraction of a second) a superposition of macroscopically different states.

 

[bhobba]

It seems that decoherence is present all the time.

Indeed it is. A state with exact position would spread over time to become inexact. Its constant decoherence that stops that.

Thanks
Bill

 

[kith]

Every density matrix can be decomposed into the form $\rho = \sum_j p_j |\psi_j\rangle \langle \psi_j |$. Once you've done this decomposition, it can be interpreted as "The system is in one of the pure states $|\psi_j\rangle$, but we don't know which."

Yes, in fact this is how the density matrix is often introduced. But there seems to be something odd about it to me.

Suppose you create a proper mixture by mixing pure states. The resulting density matrix has the straightforward interpretation which you gave above.

Now let's change to a new set of state vectors $\{ |\phi_i\rangle \}$ which is disjoint with the old set and also satisfies $\rho = \sum_i p'_i |\phi_i\rangle \langle \phi_i |$. Don't we now know for sure that our system isn't in any of these new states because we know that it is in one of the states we used for the mixing?

 

[stevendaryl]

Yes, in fact this is how the density matrix is often introduced. But there seems to be something odd about it to me.

Suppose you create a proper mixture by mixing pure states. The resulting density matrix has the straightforward interpretation which you gave above.

Now let's change to a new set of state vectors $\{ |\phi_i\rangle \}$ which is disjoint with the old set and also satisfies $\rho = \sum_i p'_i |\phi_i\rangle \langle \phi_i |$. Don't we now know for sure that our system isn't in any of these new states because we know that it is in one of the states we used for the mixing?

Well, that won't work with absolutely every possible basis. In general, if you pick a basis $|\phi_j\rangle$, then in terms of that basis, the density matrix will look like: $\rho = \sum_{i j} C_{ij} |\phi_j\rangle \langle \phi_i |$. Only for certain choices of basis will it be true that $C_{ij} = 0$ when $i \neq j$.

But you're right, there will be more than one way to write a density matrix as a mixture of pure states, which calls into question the idea that it's really in some pure state, we just don't know which. My point is that if you just pick one way of writing it as a mixture of pure states and assume that it's a proper mixture, you won't run into any contradictions. That doesn't mean that it's true, only that it's safe to pretend it's true.

 

[naima]

I think I can explain Atyy's quote using a pair of entangled electrons:

I understand now why i had a problem with atyy's sentences. the problem for me was in the words "system" and "subsysrem".
He first says that "In an improper mixed state, each system is in the same pure state" he speaks of the composite global system system described by a ray in a Hilbert space. The next sentence begins like this:
"The improper mixed state is the state of a subsystem".
I hope that Atyy will be here soon.

 

[kith]

My point is that if you just pick one way of writing it as a mixture of pure states and assume that it's a proper mixture, you won't run into any contradictions. That doesn't mean that it's true, only that it's safe to pretend it's true.

I'm struggling a bit with circularity here. What you say doesn't work for reduced density matrices when the combined system is in an entangled state, does it? Because in case it did, Bell tests would be analogous to Bertelsmann's socks. So what you seem to be saying is that if we have a mixture which is not improper, we can assume it to be proper?

 

[A. Neumaier]

My point is that if you just pick one way of writing it as a mixture of pure states and assume that it's a proper mixture, you won't run into any contradictions. That doesn't mean that it's true, only that it's safe to pretend it's true.

And different people will pretend different things. This is a very bad way of introducing subjectiveness into interpretations.

 

[stevendaryl]

I'm struggling a bit with circularity here. What you say doesn't work for reduced density matrices when the combined system is in an entangled state, does it? Because in case it did, Bell tests would be analogous to Bertelsmann's socks. So what you seem to be saying is that if we have a mixture which is not improper, we can assume it to be proper?

In the case of an entangled pair of particles, you CAN'T write the density matrix in the form $\sum_i p_i |\psi_i\rangle \langle \psi_i|$. The state of an entangled pair is $\frac{1}{\sqrt{2}}(|U, D \rangle - |D,U\rangle)$. If you form the density matrix, you get:

$\frac{1}{2}(|U,D\rangle \langle U, D| - |U,D\rangle \langle D, U| - |D, U\rangle\langle U, D| + |D,U\rangle\langle D,U|$

 

[kith]

In the case of an entangled pair of particles, you CAN'T write the density matrix in the form $\sum_i p_i |\psi_i\rangle \langle \psi_i|$.

I was talking about the reduced density matrices of the individual particles which can be written in this form.

 

[stevendaryl]

And different people will pretend different things. This is a very bad way of introducing subjectiveness into interpretations.

The question was about how decoherence can mimic the effects of wave function collapse. I tried to answer that question. Different people find different sorts of explanations helpful.

 

[stevendaryl]

I was talking about the reduced density matrices of the individual particles which can be written in this form.

You get a reduced matrix when you trace out unobservable degrees of freedom. Both particles are observable in EPR, so tracing out one of the particles to get a reduced matrix for the other particle is throwing away information.

 

[kith]

Both particles are observable in EPR, so tracing out one of the particles to get a reduced matrix for the other particle is throwing away information.

Agreed. But then what are your objections to my post #46? My point was very similar to yours: that we cannot interpret the reduced density matrix of an entangled state as a proper mixture. So your answer to my question there seems to be yes.

 

[A. Neumaier]

In the case of an entangled pair of particles, you CAN'T write the density matrix in the form $\sum_i p_i |\psi_i\rangle \langle \psi_i|$. The state of an entangled pair is $\frac{1}{\sqrt{2}}(|U, D \rangle - |D,U\rangle)$. If you form the density matrix, you get:
$\frac{1}{2}(|U,D\rangle \langle U, D| - |U,D\rangle \langle D, U| - |D, U\rangle\langle U, D| + |D,U\rangle\langle D,U|$

Every density matrix $\rho$, and hence in particular yours, can be written in the stated form, by taking the $\psi_i$ as an orthonormal basis of eigenvectors of $\rho$. One gets mixed terms only if one insists on choosing a particular basis. In the case of a pure state, as in your example, the sum consists of a single term only.

 

[Ralph Dratman]

[This link] explains it clearly:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf
It is however technical. To understand it you must learn the technicalities.

That is an excellent paper which delivers on the promise of a clear explanation.